Angular Acceleration differentation w.r.t to time

Hello everyone,

I want to derivate my simple function w.r.t time to get the acceleration and therefore I use the diff syntax.

I define my variable with q=omega*t but then omega the angular speed is constant.

If I derivate my equation twice with respect to time, I don´t have an angular_acceleration in my acceleration equation. The analytical equation of the acceleration should be f_dot_dot = cos(omega*t)*angular_acceleration - sin(omega*t)*omega^2.

What can I do to get an angular_acceleration?

t = 0:0.01:2*pi; % Time Interval
omega = 2; % Angular speed
q = omega*t; % Angular q

f = sin(q); % Function

f_dot = diff(f)./diff(t); % First derivative w.r.t
f_dot = [f_dot, NaN]; % Vector has the same length as at the beginning
f_dot_dot = diff(f_dot)./diff(t); % Second derivative w.r.t

Best regards

Matthias

[darova]

f_dot = diff(f)./diff(t);
f_dot_dot = diff(f_dot)./diff(t(2:end));
    % or
f_dot_dot  = diff(f)./diff(t(2:end)).^2;

[mattzoe]

Hello darova,

thanks for your help, but in my case I need the angular acceleration and I dont´t know how do define it.

Best regards

[darova]

If the velocity is a constant and acceleration is a change of velocity? How can you have acceleration?

[mattzoe]

@darvora that's right. But I don't know how to define my equation to solve it numerically with an included angular acceleration.

I could use the symbolic differentiation to get an angular acceleration (q_dot_dot), but I would prefer the upper one.

syms q(t)
f = sin(q);
f_dot_dot = diff(f,t,2);

f_dot_dot = cos(q)*q_dot_dot - sin(q)*q_dot^2.

Best regards

Answer 1

[SHABNUR MANSURI]

t = 0:0.01:2*pi; % Time Interval

omega = 2; % Angular speed

q = omega*t; % Angular q

f = sin(q); % Function

f_dot = diff(f)./diff(t); % First derivative w.r.t

f_dot = [f_dot, NaN]; % Vector has the same length as at the beginning

f_dot_dot = diff(f_dot)./diff(t); % Second derivative w.r.t